Q.A uniform circular plate of radius (R) has a smaller circular portion of radius (R/2) cut out from its edge. Determine the position of the center of mass of the remaining portion of the plate with respect to the center of the original plate.
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What is the Center of Mass?
Imagine you pick up a broom by the handle and try to balance it horizontally on one finger. You instinctively slide your finger along the handle until the broom stays level. That point — the one where the broom doesn't tip — is its center of mass.
Now think about throwing a cricket bat. It spins and wobbles in the air, but there is one point on the bat that follows a smooth, parabolic path, as if all the bat's mass were concentrated there. That point is also the center of mass.
The core idea is simple: the center of mass is the average position of all the mass in an object. It's the point where you could imagine the entire mass of the object being concentrated, and the object would behave the same way under the influence of external forces.
Why does this matter?
When you push an object at its center of mass, it moves in a straight line without rotating. Push it anywhere else, and it will both move and spin. This is why:
- A car's stability depends on where its center of mass is (lower = safer).
- A tightrope walker holds a long pole — moving the pole shifts their combined center of mass back over the rope.
- In projectile motion, the center of mass of a system (like an exploding firework) continues along the original parabolic path, even though the fragments scatter.
The precise definition
For a system of particles, the center of mass is the weighted average of their positions, where the weight is the mass of each particle.
RCM=m1+m2+⋯+mnm1r1+m2r2+⋯+mnrn=∑mi∑miri
Here:
- RCM is the position vector of the center of mass
- mi is the mass of the i-th particle
- ri is the position vector of that particle
For a continuous object (like a rod or a sphere), the sum becomes an integral:
RCM=M1∫rdm
where M is the total mass and dm is an infinitesimal mass element.
Breaking it down with an example
Take two masses on a light rod: m1=2 kg at x=0, and m2=3 kg at x=5 m.
The center of mass is:
xCM=2+3(2)(0)+(3)(5)=50+15=3 m
So the center of mass is at x=3 m, closer to the heavier mass. That makes intuitive sense — the heavier mass "pulls" the average toward itself.
The center of mass does not have to be inside the object. A ring or a hollow sphere has its center of mass at the geometric center, which is empty space.
Key properties to remember
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External forces only — Internal forces (like collisions between parts of the system) do not affect the motion of the center of mass. Only external forces can change its velocity.
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If no external force acts, the center of mass moves with constant velocity (or stays at rest). This is the law of conservation of momentum applied to the whole system.
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For symmetric objects with uniform density, the center of mass coincides with the geometric center. For irregular shapes, it shifts toward the region with more mass. …
Since the cut-out piece's own center of mass sits off-center, removing it shifts the center of mass of what remains to the opposite side. …
The center of mass of the remaining portion shifts by R/6 from the center of the original plate, on the side opposite to the cut-out.
Let the original uniform circular plate have radius R, mass M, and surface mass density σ = M/(πR²), centered at the origin O.
A smaller circular portion of radius R/2 is cut out from the edge — meaning its own edge is tangent to the boundary of the original plate, so its center lies at a distance R/2 from O (call this point along the +x axis, at x = R/2).
Mass of the original full plate: m₁ = σπR² = M (at position x = 0).
Mass of the cut-out piece: m₂ = σπ(R/2)² = σπR²/4 = M/4 (at position x = R/2).
The center of mass of the (full plate) can be thought of as: [full plate] = [remaining portion] + [cut-out piece]. Since the full plate's COM is at the origin:
m₁·(0) = m_remaining·x_cm + m₂·(R/2)
where m_remaining = m₁ − m₂ = M − M/4 = 3M/4.
Solving for x_cm:
0 = (3M/4)·x_cm + (M/4)(R/2) …
Showing the 12 most recent of 22 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.In the absence of an external force, velocity of the center of mass -(a) Remains constant(b) Decreases(c) is Zero(d) Increases
›Reveal solutionSolution
In the absence of an external force, the velocity of the centre of mass of a system remains constant (Newton's first law applied to the system as a whole).
For a system of particles, the equation of motion of the centre of mass is F_ext = M (d v_cm/dt), where M is the total mass and F_ext is the net external force. This follows because all internal forces (between particles of the system) occur in Newton's-third-law pairs and cancel out when summed over the whole system. If F_ext = 0, then d v_cm/dt = 0, meaning the centre of mass m …
- CBSE 2026Set ANNUAL1 markMCQQ.There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are:(a) (2, -1)(b) (8/3, -1/3)(c) (-1/3, 8/3)(d) None of the above
›Reveal solutionSolution
xcm=m1+m2m1x1+m2x2, ycm=m1+m2m1y1+m2y2 give (−31,38).
For a system of point masses, the centre of mass is:
xcm=m1+m2m1x1+m2x2,ycm=m1+m2m1y1+m2y2
With m1=1 kg at (1,2) and m2=2 kg at (−1,3):
xcm=1+21(1)+2(−1)=31−2=−31 …
- CBSE 2026Set ANNUAL1 markMCQQ.If a uniform rod has length l, then its centre of mass will be located at(a) l/2(b) 3l/4(c) l/3(d) none of these
›Reveal solutionSolution
A uniform rod's centre of mass is at its midpoint, l/2. Answer (A).
For a uniform rod the linear mass density is constant, so the mass is symmetric about the middle of the rod. By symmetry the centre …
- CBSE 2026Set ANNUAL1 markMCQQ.If the centre of gravity of a body coincides with its centre of mass, then the gravitational fields acting on different parts of the body are(a) zero(b) equal(c) different(d) none of these
›Reveal solutionSolution
Centre of gravity coincides with centre of mass when the gravitational field is uniform (equal on all parts). Answer (B).
The centre of mass depends only on the mass distribution. The centre of gravity is the point where the total weight appears to act, which depends on how g varies over the body.
…
- CBSE 2026Set ANN1 markMCQQ.For which of the following shapes, the centre of mass does not lie on the body?(a) Pencil(b) Dice(c) Shotput(d) Bangle
›Reveal solutionSolution
A bangle's centre of mass lies at its geometric centre - empty space - so the centre of mass does not lie on the body.
The centre of mass of a body need not lie within the material of the body; it depends on how the mass is distributed. For symmetric bodies it lies at the geometric centre.
- Pencil: centre of mass lies on the pencil (its midpoint).
- Dice (cube): centre of mass at the solid centre, inside the body. …
- CBSE 2025Set ANNUAL1 markMCQQ.Where is the centre of mass of a uniform rod located?(a) At one end of the rod(b) At the midpoint of the rod(c) At the top of the rod(d) At the bottom of the rod
›Reveal solutionSolution
For any uniformly (symmetrically) shaped and uniformly dense body like a straight rod, the centre of mass lies at the geometric centre -- the midpoint.
Centre of mass, x_cm = (Integral of x dm) / (Integral of dm), over the length of the rod.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Case study: The 'centre of mass' of a body is a point at which whole of the mass of the body is supposed to be concentrated. The centre of mass of a two body system is- X_cm = (m1 x1 + m2 x2) / (m1 + m2) where symbols have their usual meaning. The velocity of centre of mass is V_cm = dX_cm / dt.(1) Which of the following statement is false?(a) Centre of mass of body may lie outside the body.(b) Centre of mass of body may lie on the body.(c) Centre of mass of two bodies of different masses lie at the centre of line joining the masses.(d) Centre of mass of a ring lies at a point where actually no mass of the body exists.
›Reveal solutionSolution
Statement (c) is the false one — the COM of two unequal masses lies closer to the heavier mass, not at the geometric midpoint.
The centre of mass of a two-particle system lying on a line, with masses m1 and m2 at positions x1 and x2, is given by:
Xcm=m1+m2m1x1+m2x2
Check each statement:
- 'Centre of mass of body may lie outside the body' — TRUE. Example: a ring, a horseshoe, or a boomerang — the COM lies in the hollow/empty region, outside the actual material of the body.
- 'Centre of mass of body may lie on the body' — TRUE. For a uniform, solid, symmetric body like a sphere or cube, the COM lies at the geometric centre, which is a point within the body's material. …
- CBSE 2025Set ANNUAL1 markMCQQ.(2) If two particle of equal mass are placed at a distance 'd' apart, the centre of mass is located(a) At one of the particles.(b) At the midpoint of the distance between them.(c) Outside the line joining the two particles.(d) None of above
›Reveal solutionSolution
Two equal masses have their centre of mass exactly at the midpoint of the line joining them.
Let the two particles, each of mass m, be placed at positions x1=0 and x2=d on a line. Using the centre-of-mass formula:
Xcm=m1+m2m1x1+m2x2=m+mm(0)+m(d)=2mmd=2d
…
- CBSE 2025Set ANNUAL1 markMCQQ.(3) The motion of centre of mass of a system is determined by(a) External forces only(b) Internal forces only(c) Both internal and external forces(d) Number of particles in the system
›Reveal solutionSolution
Only external forces determine the motion of the centre of mass; internal forces always cancel by Newton's third law.
For a system of particles, the total mass M times the acceleration of the centre of mass equals the sum of ALL forces acting on all the particles:
Macm=∑Fexternal+∑Finternal
By Newton's third law, every internal force (a particle of the system exerting a force on another particle of the same system) comes as an action-reaction pair that is equal in magnitude and opposite in direction. When summed over the whole system, all such internal force pairs cancel exactly:
∑Finternal=0
So the equation of motion for the centre of mass simplifies to:
Macm=∑Fexternal
…
- CBSE 2025Set ANNUAL1 markMCQQ.(4) A body moving in a straight line explodes in 20 pieces. The path of centre of mass will be(a) Parabolic(b) Straight line(c) Circular(d) Hyperbolic
›Reveal solutionSolution
The COM keeps moving along the original straight-line path, since explosion forces are purely internal.
As derived from Newton's laws applied to a system of particles, the acceleration of the centre of mass of a system is determined ONLY by the net EXTERNAL force acting on it; internal forces (however violent) always cancel in pairs by Newton's third law and cannot change the COM's motion.
When a body moving in a straight line explodes into 20 pieces, the forces of the explosion (which fling the pieces apart) are entirely internal to the system made up of all 20 pieces together. No new external force is introduced by the explosion itself (only gravity, if present, would be external — and the question is framed purely on the explosion's effect).
…
- CBSE 2024Set ANNUAL1 markMCQQ.Out of two particles of same mass, one is stationary and acceleration of another is a⃗. The acceleration of centre of mass of the system will be (A) zero (B) a⃗/2 (C) a⃗ (D) 2a⃗
›Reveal solutionSolution
Centre-of-mass acceleration of the two-particle system is a/2.
For a system of particles, acm=∑mi∑miai. Here both particles have equal mass m: one is stationary (a1=0) and the other has acceleration a …
- CBSE 2024Set ANNUAL1 markMCQQ.If a stationary firecracker explodes into a number of particles, then the centre of mass will(a) move in vertical direction(b) move in concentric circles(c) remain stationary(d) move in parabolic path.
›Reveal solutionSolution
Internal explosive forces cannot move the centre of mass; only an external force can, so it stays at rest.
The forces that blow the firecracker into fragments are internal forces (action-reaction pairs between the fragments) — by Newton's third law they always occur in equal and opposite pairs and cancel out when summed over the whole system. The motion of the centre of mass of a system is governed only by the net external force acting on it: Macm=Fext. Since gravity is the only external force here (assume negligible over the short duration, and the cracker was stationary be …
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